The first-order rate constant for the decomposition of CH3N₂CH3, CH3N₂CH3(g) → C₂H5(9) + N₂(g) at 327 °C is k = 3.60e- 04 s ¹. Suppose we start with 0.0858 mol of CH3N₂CH3 in a volume of 6.9 L. How many moles of CH3N₂CH3 will remain after 3.79e+01 min? a) 0.0846 moles will remain. b) 0.00117 moles will remain. c) 6.81e-79 moles will remain. d) 0.0703 moles will remain. e) 0.0378 moles will remain.
The first-order rate constant for the decomposition of CH3N₂CH3, CH3N₂CH3(g) → C₂H5(9) + N₂(g) at 327 °C is k = 3.60e- 04 s ¹. Suppose we start with 0.0858 mol of CH3N₂CH3 in a volume of 6.9 L. How many moles of CH3N₂CH3 will remain after 3.79e+01 min? a) 0.0846 moles will remain. b) 0.00117 moles will remain. c) 6.81e-79 moles will remain. d) 0.0703 moles will remain. e) 0.0378 moles will remain.
Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter12: Kinetics
Section: Chapter Questions
Problem 60E: The rate constant for the decomposition of acetaldehyde, CH3CHO, t0 methane, CH4, and carbon...
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![The first-order rate constant for the decomposition of CH3N₂CH3, CH3N₂CH3(g) → C₂H5(9) + N₂(g) at 327 °C is k = 3.60e-
04 s ¹. Suppose we start with 0.0858 mol of CH3N₂CH3 in a volume of 6.9 L. How many moles of CH3N₂CH3 will remain
after 3.79e+01 min?
a) 0.0846 moles will remain.
b) 0.00117 moles will remain.
c) 6.81e-79 moles will remain.
d) 0.0703 moles will remain.
e) 0.0378 moles will remain.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86624ab7-da2a-4a5b-91b8-0bf4ad8ab366%2F5edd2729-2557-4ef0-a849-8f3b6dcdf242%2Fwts43rl_processed.png&w=3840&q=75)
Transcribed Image Text:The first-order rate constant for the decomposition of CH3N₂CH3, CH3N₂CH3(g) → C₂H5(9) + N₂(g) at 327 °C is k = 3.60e-
04 s ¹. Suppose we start with 0.0858 mol of CH3N₂CH3 in a volume of 6.9 L. How many moles of CH3N₂CH3 will remain
after 3.79e+01 min?
a) 0.0846 moles will remain.
b) 0.00117 moles will remain.
c) 6.81e-79 moles will remain.
d) 0.0703 moles will remain.
e) 0.0378 moles will remain.
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